Solve the equation. $\dfrac{dy}{dx}=xe^{-y}+10e^{-y}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\dfrac{x^2}{2}+10x+C$ (Choice B) B $y=-\ln\left(-\dfrac{x^2}{2}-10x+C\right)$ (Choice C) C $y=\ln\left(\dfrac{x^2}{2}+10x+C\right)$ (Choice D) D $y=-\ln\left(-\dfrac{x^2}{2}-10x\right)+C$
Explanation: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=xe^{-y}+10e^{-y} \\\\ \dfrac{dy}{dx}&=\dfrac{x+10}{e^y} \\\\ e^y\,dy&=(x+10)\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} e^y\,dy&=(x+10)\,dx \\\\ \int e^y\,dy&=\int (x+10)\,dx \\\\ e^y&=\dfrac{x^2}{2}+10x+C \\\\ \ln(e^y)&=\ln\left(\dfrac{x^2}{2}+10x+C\right) \\\\ y&=\ln\left(\dfrac{x^2}{2}+10x+C\right) \end{aligned}$ Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\ln\left(\dfrac{x^2}{2}+10x+C\right)$